Question

Complete the following

Mn²⁺(aq) + F₂(g) --> F^-(aq) + MnO₄^-(aq)

a. Balance the redox equation above in acidic solution. Show work for full credit.

b. Circle the reducing agent in this redox reaction.

c. Calculate E^o for this reaction.

Answer 1

half reaction is Mn2+ ---> MnO4-    , to balance O we add H2O

Mn2+ + 4H2O ---> MnO4- , to balance H we add H+

Mn2+ + 4H2O -----> MnO4- + 8H+ , to balance charge we add e-

Mn2+(aq) + 4H2O (l) ----> MnO4^-(aq) + 8H+(aq) + 5e- (aq) ..................(1)

second half is F2(g) ---> 2F-   we balance charge by adding e-

F2(g) + 2e-(aq) ----> 2F-(aq) ..................(2)

we add both equations such that total electron cancell each other

now reaction (1) is multiplied with 2 and added to 5 times reaction (2)

2Mn2+(aq) + 5F2(g) +8H2O (l) ----> 2MnO4^-1(aq) + 16H+ (aq) + 10F-(aq)

Reducing agent is one which gets oxidised , hence it is Mn2+    ( as Mn2+ gets oxdised to Mn+7)

Eo rxn = Eo( F2/f-) -Eo(MnO4-,H+/M2+)

        = 2.87 - 1.51   = 1.36